Fundamentele Informatica 1 (I&E) najaar 2014 - - PowerPoint PPT Presentation

Fundamentele Informatica 1 (I&E)

najaar 2014 http://www.liacs.leidenuniv.nl/~vlietrvan1/fi1ie/ Rudy van Vliet rvvliet(at)liacs(dot)nl college 11, maandag 1 december 2014

• 6. Context-Free and Non-Context-Free Languages

6.1. The Pumping Lemma for Context-Free Languages

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• 6. Context-Free

and Non-Context-Free Languages 6.1. The Pumping Lemma for Context-Free Languages

• reg. languages

FA

• reg. grammar
• reg. expression
• determ. cf. languages

DPDA

• cf. languages

PDA

• cf. grammar
• re. languages

TM

• unrestr. grammar

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A slide from lecture 3:

2.4 The Pumping Lemma

Theorem 2.29. The Pumping Lemma for Regular Languages. Suppose L is a language over the alphabet Σ. If L is accepted by a finite automaton M = (Q, Σ, q0, A, δ), and if n is the number of states of M, then for every x ∈ L satisfying |x| ≥ n, there are three strings u, v, and w such that x = uvw and the following three conditions are true:

• 1. |uv| ≤ n.
• 2. |v| > 0 (i.e., v = Λ).
• 3. For every i ≥ 0, the string uviw also belongs to L.

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A slide from lecture 3: Example 2.30. The language AnBn. Let L = {aibi | i ≥ 0}.

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Now, context-free languages. Intuitively clear that PDA cannot accept AnBnCn or XX. . .

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Now, context-free languages. Intuitively clear that PDA cannot accept AnBnCn or XX. . . Pumping lemma based on derivation in CFG (not on PDA): S ⇒∗ vAz ⇒∗ v wAy z ⇒∗ vw x yz S ⇒∗ vAz ⇒∗ v wAy z ⇒∗ vw wAy yz ⇒∗ vwmxymz

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Theorem 6.1. The Pumping Lemma for Context-Free Languages. Suppose L is a context-free language. Then there is an integer n so that for every u ∈ L with |u| ≥ n, u can be written as u = vwxyz, for some strings v, w, x, y and z satisfying

• 1. |wy| > 0
• 2. |wxy| ≤ n
• 3. for every m ≥ 0, vwmxymz ∈ L
• Proof. . .

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A slide from lecture 7: Definition 4.29. Chomsky Normal Form A context-free grammar is said to be in Chomsky normal form if every production is of one of these two types: A → BC (where B and C are variables) A → σ (were σ is a terminal symbol)

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A slide from lecture 8: Theorem 4.30. (not Theorem 4.31!) For every context-free grammar G, there is another CFG G1 in Chomsky normal form such that L(G1) = L(G) − {Λ}. What if Λ / ∈ L(G) ?

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Number of leaf nodes in a binary tree of a given height

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Theorem 6.1. The Pumping Lemma for Context-Free Languages. Proof Let G be CFG in Chomsky normal form with L(G) = L − {Λ}. Derivation tree in G is binary tree (where each parent of a leaf node has only one child). Height of a tree is number of edges in longest path from root to leaf node. At most 2h leaf nodes in binary tree of height h: |u| ≤ 2h.

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At most 2h leaf nodes in binary tree of height h: |u| ≤ 2h. Let p be number of variables in G, let n = 2p and let u ∈ L(G) with |u| ≥ n. (Internal part of) derivation tree of u in G has height at least p. Hence, longest path in (internal part of) tree contains at least p + 1 (internal) nodes. Consider final portion of longest path in derivation tree. (leaf node + p + 1 internal nodes), with ≥ 2 occurrences of a variable A. Pump up derivation tree, and hence u.

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Pumping Lemma for CFLs

✖✕ ✗✔ ✁ ✁ ❅ ❅ ❅

S

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆

v z

✖✕ ✗✔ ❆ ❆

• A

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆

w y

✖✕ ✗✔

A

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆

x

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Pumping Lemma for CFLs

✖✕ ✗✔ ✁ ✁ ❅ ❅ ❅

S

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆

v z

✖✕ ✗✔

A

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆

x

✖✕ ✗✔ ✁ ✁ ❅ ❅ ❅

S

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆

v z

✖✕ ✗✔ ❆ ❆

• A

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆

w y

✖✕ ✗✔ ❆ ❆

• A

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆

w y

✖✕ ✗✔

A

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆

x

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Application of pumping lemma: mainly to prove that a language L cannot be generated by a context-free grammar. How? Find a string u ∈ L with |u| ≥ n that cannot be pumped up! What is n? What should u be? What can v, w, x, y and z be?

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Suppose that there exists context-free grammar G with L(G) = L. Let n be the integer from the pumping lemma. Pumping lemma: For every u ∈ L with |u| ≥ n, there are five strings v, w, x, y and z such that u = vwxyz and the following three conditions are true:

• 1. |wy| > 0
• 2. |wxy| ≤ n
• 3. for every m ≥ 0, vwmxymz ∈ L

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Suppose that there exists context-free grammar G with L(G) = L. Let n be the integer from the pumping lemma. We prove: NOT ( For every u ∈ L with |u| ≥ n, there are five strings v, w, x, y and z such that u = vwxyz and the following three conditions are true:

• 1. |wy| > 0
• 2. |wxy| ≤ n
• 3. for every m ≥ 0, vwmxymz ∈ L

)

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Suppose that there exists context-free grammar G with L(G) = L. Let n be the integer from the pumping lemma. We prove: There exists u ∈ L with |u| ≥ n, such that NOT ( there are five strings v, w, x, y and z such that u = vwxyz and the following three conditions are true:

• 1. |wy| > 0
• 2. |wxy| ≤ n
• 3. for every m ≥ 0, vwmxymz ∈ L

)

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Suppose that there exists context-free grammar G with L(G) = L. Let n be the integer from the pumping lemma. We prove: There exists u ∈ L with |u| ≥ n, such that for every five strings v, w, x, y and z such that u = vwxyz NOT ( the following three conditions are true:

• 1. |wy| > 0
• 2. |wxy| ≤ n
• 3. for every m ≥ 0, vwmxymz ∈ L

)

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Suppose that there exists context-free grammar G with L(G) = L. Let n be the integer from the pumping lemma. We prove: There exists u ∈ L with |u| ≥ n, such that for every five strings v, w, x, y and z such that u = vwxyz NOT all of the following three conditions are true:

• 1. |wy| > 0
• 2. |wxy| ≤ n
• 3. for every m ≥ 0, vwmxymz ∈ L

)

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Suppose that there exists context-free grammar G with L(G) = L. Let n be the integer from the pumping lemma. We prove: There exists u ∈ L with |u| ≥ n, such that for every five strings v, w, x, y and z such that u = vwxyz if

• 1. |wy| > 0
• 2. |wxy| ≤ n

then NOT (

• 3. for every m ≥ 0, vwmxymz ∈ L

)

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Suppose that there exists context-free grammar G with L(G) = L. Let n be the integer from the pumping lemma. We prove: There exists u ∈ L with |u| ≥ n, such that for every five strings v, w, x, y and z such that u = vwxyz if

• 1. |wy| > 0
• 2. |wxy| ≤ n

then

• 3. there exists m ≥ 0, such that vwmxymz does not belong to L

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Example 6.3. Applying the Pumping Lemma to AnBnCn AnBnCn = {aibici | i ≥ 0} Choose u = . . .

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Example 6.3. Applying the Pumping Lemma to AnBnCn AnBnCn = {aibici | i ≥ 0} Choose u = anbncn

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A slide from lecture 3:

2.4 The Pumping Lemma

Theorem 2.29. The Pumping Lemma for Regular Languages. Suppose L is a language over the alphabet Σ. If L is accepted by a finite automaton M = (Q, Σ, q0, A, δ), and if n is the number of states of M, then for every x ∈ L satisfying |x| ≥ n, there are three strings u, v, and w such that x = uvw and the following three conditions are true:

• 1. |uv| ≤ n.
• 2. |v| > 0 (i.e., v = Λ).
• 3. For every i ≥ 0, the string uviw also belongs to L.

25

Theorem 6.1. The Pumping Lemma for Context-Free Languages. Suppose L is a context-free language. Then there is an integer n so that for every u ∈ L with |u| ≥ n, u can be written as u = vwxyz, for some strings v, w, x, y and z satisfying

• 1. |wy| > 0
• 2. |wxy| ≤ n
• 3. for every m ≥ 0, vwmxymz ∈ L
• Proof. . .

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Example 6.5. Applying the Pumping Lemma to. . . {x ∈ {a, b, c}∗ | na(x) < nb(x) and na(x) < nc(x)} Choose u = . . .

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Example 6.5. Applying the Pumping Lemma to. . . {x ∈ {a, b, c}∗ | na(x) < nb(x) and na(x) < nc(x)} Choose u = anbn+1cn+1

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Example 6.6. The Set of Legal C Programs is Not a CFL Choose u = . . .

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Example 6.6. The Set of Legal C Programs is Not a CFL Choose u = main(){int aaa...a;aaa...a;aaa...a;} where aaa...a contains n + 1 a’s

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Example 6.4. Appplying the Pumping Lemma to XX XX = {xx | x ∈ {a, b}∗} Choose u = . . .

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Example 6.4. Appplying the Pumping Lemma to XX XX = {xx | x ∈ {a, b}∗} Choose u = anbnanbn

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